Edited
I'm reading Shiryaev's Probability and turns out that he doesn't provide an example of how do I construct the random variable $E(\xi|G)$ where $\xi$ is an RV and $G$ is a $\sigma$-algebra. The definition is the following: The conditional expectation fo an RV $\xi$ wrt to a $\sigma$-algebra $G$ is the RV denoted by $E(\xi|G)$ meeting the criteria: 1) $E(\xi|G)$ is $G$-mesurable and 2) for all $A \in G$, holds that
$$ \int_A \xi d\mathbb{P} = \int_A E(\xi|G) d\mathbb{P}. $$
I could find and understand an example for the case where $\xi$ is discrete and $G$ is discrete and finite, but I'm having trouble to compute it by defintion when $\xi$ is continuous.
Searching the internet, I've found this example (page 43).
The example considers $\Omega=(0,1]$, the Lebesgue measure and the Borel sigma-algebra over $\Omega$ and considers $X(\omega)$ a random variable, Borel measurable. Also $G$ is the $\sigma$-algebra generated by intervals $\left(\dfrac{j-1}{n},\dfrac{j}{n}\right], j=1,2,\ldots$. The example then states that $E(X|G)$ is given by:
$$ E(X|G)(\omega) = \int_{(j-1)/n}^{j/n} X(s)ds \text{ if } \omega \in \left(\dfrac{j-1}{n},\dfrac{j}{n}\right] $$ and I can't understand why this holds. More specifically: let's assume $X(\omega)=\omega^2$. Under the same conditions of the example, how do I compute (from definition) the random variable $E(X|G)$? I'm trying to get the "step-by-step" solutions for this kind of stuff, because to me the definition is a bit blurry/abstract and books I read or consulted provide no example for that.
Now Imagine I'm dealing with the same example, but instead I'm working with the smallest $\sigma$-algebra generated by a set that is non-enumerable, say: $G= \sigma(\{\omega \in \Omega: 0 < \Omega \leq 0.5\})$ (or any other endpoints that are lie within $\Omega$. In that case, is it possible to write explicitly the random variable $E(\xi|G)$? What are the "minimum conditions" that allow me to write this RV explicitly, using the definition?
Thanks!!
Best Answer
The sigma algebra generated by $(\frac {j-1} n, \frac j n], 1\leq j \leq n$ is simply the collection of all possible unions of these intervals. [This fact is true for sigma algebra generated by any countable partition of $\Omega$]. To verify that the formula given in the example for $E(X|G)$ works you only have to verify that $\int_A XdP=\int E(X|G)dP$ when $A$ is one of these intervals (because, then, it will hold for all possible unions of these intervals also). But when $A$ is one of these intervals the verification is very simple.
If $X(\omega)=\omega^{2}$ then $E(X|G)$ is the r.v. which has the constant value $\frac 1 3((\frac j n)^{2}-(\frac {j-1} n)^{2})$ on the interval $(\frac {j-1} n, \frac j n]$ for $i \leq j \leq n$.